Middle-题目36：74. Search a 2D Matrix

题目原文：
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.
For example,

Consider the following matrix:

[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
题目大意：
在一个满足下列性质的矩阵中寻找一个数：
每一行从左到右，每一列从上到下的数组是递增顺序的。且下一行最小的数比上一行最大的数大。
题目分析：
方法一：既然做过了Middle-题目32，就可以用同样的算法。
方法二：因为在第32题的基础上又加一个限定，所以对每行第一个数二分查找，确定该数可能存在的行，再在行内二分查找。
源码：（language：java）
方法一略。
方法二：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int start = 0,end = matrix.length - 1,mid;
        if(start == end) // the matrix has only 1 line
            return binarySeach(matrix[0],target);
        while(start <= end) {
            mid = (start + end) / 2;
            if(matrix[mid][0] == target) 
                return true;
            else if(matrix[mid][0] < target && (mid == matrix.length-1 || matrix[mid+1][0] > target)) // target may be at line of mid
                return binarySeach(matrix[mid],target);
            else if(matrix[mid+1][0] <= target)   //target lies after line of mid
                start = mid + 1;
            else
                end = mid-1;
        }
        return false;
    }
    private  boolean binarySeach(int[] nums, int target) {
        int start = 0, end = nums.length-1, mid;
        while(start <= end) {
            mid = (start + end) / 2;
            if(nums[mid] == target)
                return true;
            else if(nums[mid] < target)
                start = mid + 1;
            else
                end = mid - 1;
        }
        return false;
    }
}

成绩：
方法一：2ms，beats 3.01%，众数1ms,89.18%
方法二：1ms，beats 6.15%